Równanie trygonometryczne.

[MiedzianyDawid]

Zad.1
Rozwiązać równanie \(\sin \frac{x}{2}\)+\(\cos \frac{x}{2}\)=\(\sqrt{2} \sin x\)

[panb]

\(\sin \frac{x}{2}+\cos\frac{x}{2}=\sqrt2\sin x \qquad \qquad \,\,\, \text{wzór: } \sin \left( \frac{\pi}{2} -x\right)=\cos x \\
\sin\frac{x}{2}+\sin \left( \frac{\pi}{2}- \frac{x}{2} \right)=\sqrt2\sin x \qquad \text{wzór: } \sin x+\sin y=2\sin \frac{x+y}{2}\cos \frac{x-y}{2}\\
2\sin \frac{\pi}{4} \cos \left( \frac{x}{2} - \frac{\pi}{4} \right)=\sqrt2\sin x\\
\cos \left( \frac{x}{2} - \frac{\pi}{4} \right)=\sin x \\
\sin \left[ \frac{\pi}{2}- \left( \frac{x}{2} - \frac{\pi}{4} \right) \right]=\sin x\\
\sin \left( \frac{3}{4}\pi - \frac{x}{2} \right)=\sin x\\
\frac{3}{4}\pi - \frac{x}{2} = x-2k\pi \vee \frac{3}{4}\pi - \frac{x}{2} =\pi-x+2k\pi\\
x= \frac{\pi}{2}+2k\pi \,\,\, \vee \,\,\, x= \frac{\pi}{2}+ \frac{4}{3}k\pi\)